CJC 2025 Prelim P1 Solutions

9758/01/J2PRELIM/2025 [Turn Over CATHOLIC JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2 JC2 Preliminary Examination CANDIDATE NAME CLASS INDEX NUMBER MATHEMATICS 9758/01 Paper 1 02 Sep 2025 3 hours Additional Materials: Printed Answer Booklet List of Formulae (MF27) READ THESE INSTRUCTIONS FIRST Answer all the questions. Write your answers on the Printed Answer Booklet. Follow the instructions on the front cover of the answer booklet. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use an approved graphing calculator. Unsupported answers from a graphing calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphing calculator are not allowed in a question, you must present the mathematical steps using mathematical notations and not calculator commands. You must show all necessary working clearly. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 43 printed pages, including this cover page.

2 9758/01/J2PRELIM/2025 1 The diagram above shows the curve C with equation     2 2 19 x p y q r     , where p, q and r are constants. The vertices are  5, 3 and  1, 3 and the asymptotes are 3 15 2 2y x  and 3 3 .2 2y x  (a) Find the values of p, q and r. [3] (b) The curve D has equation     2 2 3 3x m y m    , where m is a positive constant. Find the range of values of m for which curves C and D do not intersect. [2] x y O (5, 3) (1, 3)

3 9758/01/J2PRELIM/2025 [Turn Over Solution: Q1 (a) Method : Method : Centre of hyperbola is the mid-point of  5, 3 and  1, 3 , i.e.  5 1 3 3, 3, 32 2                   2 2 2 2 2 9 19 1 x p y q r x p y q r                 Centre is  ,p q  Comparing 3p and 3q Centre of hyperbola is the point of intersection of the asymptotes. 3 15 (1)2 2y x   3 3 (2)2 2y x   Solving (1) & (2), 3x 3y           2 2 2 2 2 19 19 x p y q h x p y q r                 Centre is  ,p q  Comparing 3p and 3q x-distance from vertex to centre of hyperbola is 2, therefore 2 4r r   (b)           2 2 2 2 2 3 3 3 3 1 x m y m x y m         Ellipse, centre (3, 3) [Notice that the centre of the ellipse is also the centre of the hyperbola] 2 4 m m   Since m is a positive constant, 0 4m  .

4 9758/01/J2PRELIM/2025 2 The nth term of a sequence is given by 2 nu an bn c   . The first three terms of this sequence of numbers are 2, 6 and 12. (a) Find the values of a, b and c. [3] (b) Given that 1 1 1 1 1 N n nu N    , find 8 1 n nu    . [3]

5 9758/01/J2PRELIM/2025 [Turn Over Solution: Q2 (a) When 1n ,       2 1 1 1 2 2 ---- 1 u a b c a b c        When 2n  ,       2 2 2 2 6 4 2 6 ---- 2 u a b c a b c        When 3n ,       2 3 3 3 12 9 3 12 ---- 3 u a b c a b c        Using G.C., 1, 1 and 0a b c   (b) Method ①: As N  , 1 01N  . 1 1 1 n nu     7 8 1 1 1 1 1 11 1 7 1 1 8 n n n n n nu u u                   Method ②: 7 8 1 1 1 1 1 1 11 1 1 7 1 1 1 8 1 N N n n n n n nu u u N N                          As N  , 1 01N  . 8 1 1 8n nu    

6 9758/01/J2PRELIM/2025 3 A curve C has parametric equations 2 π πsin , 1 2sin , . 2 2x t y t t      (a) Sketch C, stating clearly the coordinates of the endpoints and the coordinates of the y-intercept(s). [2] (b) Find the exact cartesian equation of l, the normal to C at the point 1 , 24      . [4]

7 9758/01/J2PRELIM/2025 [Turn Over Solution: Q3 (a) (b) At 1 ,24      , Since 1 2siny t  , 1 2sin 2t  1sin 2t  π 6t  OR Since 2sinx t 2 1sin 4t  1sin 2t  π 6t  or π 6t  (rejected since y > 0) 2sinx t 1 2siny t  d 2sin cosd x t tt  d 2cosd y tt  d d 2cos 1d dd 2sin cos sin d y y t t xx t t t t    When π 6t  , d 1 21d 2 y x   (0,1) (1,3) (1,– 1 ) x y

8 9758/01/J2PRELIM/2025 Gradient of normal 1 2 Equation of normal: 1 12 2 4y x        1 17 2 8y x  or 8 4 17y x  (or any other equivalent form) OR Substitute 1 , 24      and gradient of normal 1 2 into y mx c  , 1 12 2 4 c     17 8c Equation of normal: 1 17 2 8y x 

9 9758/01/J2PRELIM/2025 [Turn Over 4 It is given that   ππ π f π 22 for 0 sin for 2 x xx x x x       and that    f πf 2x x  for all real values of x. (a) Sketch the graph of  fy x for π π 4x   . You do not need to label the coordinates of any stationary point(s). [3] (b) Find the exact area bounded by the curve  fy x , the x-axis and the lines 0x and 3 .2 πx [4]

10 9758/01/J2PRELIM/2025 Solution: Q4 (a) (b)         3π2 π 3π3π 22 π π 3π 2 π 3π 2 π sin d cos cos d 3π 3πcos π cos π cos d2 2 π sin 3ππ sin sin π2 π 1 x x x x x x x x x x                                   3π 3π 22 0 π 2 1d π 2 sin d2 π π 1 2π 1 unit f s xx x x x          u x d sind v xx  d 1d u x  cosv x x y O (0, 2) (2, 2) (, 0) (2, 0) (3, 0) (4, 0) (, 0) y = f (x)